Consider the following:
$$\begin{bmatrix}X & I \\ I & Y\end{bmatrix}=\begin{bmatrix}I & 0 \\ Y^{-1} & I\end{bmatrix}^T\begin{bmatrix}X-Y^{-1} & 0 \\ 0 & Y\end{bmatrix}\begin{bmatrix}I & 0 \\ Y^{-1} & I\end{bmatrix}$$
where
- $X,Y\in \mathbb{R}^{n\times n}\succ0$ (positive definite)
- $I\in \mathbb{R^{n\times n}}$ is the identity matrix.
- Suppose we know $$\begin{bmatrix}X & I \\ I & Y\end{bmatrix}\succeq 0$$
How to show the following: $$\textbf{rank}\begin{bmatrix}X & I \\ I & Y\end{bmatrix}=n+\textbf{rank}(X-Y^{-1})=n+\textbf{rank}(XY-I)$$
I cannot find a way to figure out why this is true.
Multiplying a matrix $A$ from the left or from the right by an invertible matrix doesn't change its rank. The matrix
$$ \begin{pmatrix} I & 0 \\ Y^{-1} & I \end{pmatrix} $$
is invertible (being an lower block triangular with invertible blocks on the diagonal) and so is its transpose and hence
$$ \operatorname{rank} \begin{pmatrix} X & I \\ I & Y \end{pmatrix} = \operatorname{rank} \begin{pmatrix} X - Y^{-1} & 0 \\ 0 & Y \end{pmatrix}. $$
The rank of a block diagonal matrix is the sum of ranks of the blocks on the diagonal and so
$$ \operatorname{rank} \begin{pmatrix} X - Y^{-1} & 0 \\ 0 & Y \end{pmatrix} = \operatorname{rank} (X - Y^{-1}) + \operatorname{rank} Y. $$
Since $Y > 0$, we have $\operatorname{rank} Y = n$ giving the first equality. Since $$\operatorname{rank} (X - Y^{-1}) = \operatorname{rank} ((X - Y^{-1})Y) = \operatorname{rank}( XY - I) $$ you get the second equality.